Thread: How to 'replicate' an array of hex values?

I have a set of 4 fixed bytes that i put in an array:

Code:

char fixed[] = "\xcd\xf1\x12\xbf";

Then I want to fill another newarray[1028] with the values of the array above.
(ie 257 X fixed[] inside newarray[1028]) and get something like:

Code:

char newarray[1028] = "\xcd\xf1\x12\xbf\xcd\xf1\x12\xbf\xcd\xf1\x12\xbf.........";

I tried with an ugly loop but I doesn't seem to work:

Code:

int i, j = 0;
char fixed[] = "\xcd\xf1\x12\xbf";
char newarray[1028];
for(i=0;i<1028;i++) {
    if(j > 3)
        j = 0;
    newarray[i] = fixed[j];
    j++;
}

Any idea how I can 'replicate' an array n times in another array?

You need to test 'j' after the assignment. Also, you could simply use memcpy in a loop.

EDIT: Actually, the code posted should work (unlike my brain at the moment).

Originally Posted by sef0

but if I do:

Code:

for(i=0;i<1028;i++) {
    printf("%02x",newarray[i]);
}

I don't get cdf112bf cdf112bf cdf112bf cdf112bf ... 257 times

How would I use memcpy() to replicate my array?

No, I expect that your compiler (like 95% of all compilers) have signed char, which means that your 0xcd and 0xf1 becomes fffffffcd and fffffff1 respectively. If you make either cast to unsigned char in the printf or make newarray an unsigned char in the first place, then I expect you get the right values.

A loop calling memcpy() for 4 bytes at a time is probably slower than the for loop suggested in the original post - and if the string is only 1028 bytes ever, then I can't for my life see any benefit in optimizing it [unless you have a 2MHz 4-bit processor].

--
Mats

--
Mats

>> I don't get cdf112bf cdf112bf cdf112bf cdf112bf ... 257 times

It would actually be 256 times, but anyway, what output are you getting?

>> How would I use memcpy() to replicate my array?

Something like:

Code:

char* duplicate( char* buffer, size_t buffer_length, char const* pattern, size_t pattern_length )
{
    for( char* ptr = buffer; buffer_length != 0; buffer_length -= pattern_length, ptr += pattern_length )
    {
        if( pattern_length > buffer_length )
            pattern_length = buffer_length;
        memcpy( ptr, pattern, pattern_length );
    }
    return buffer;
}

EDIT:
OK, so it wasn't really any simpler than the original code.

Code:

void rep( char *buf, size_t bufsiz, char *pat, size_t patsiz )
{
    size_t x = 0, y = 0;
    while( x < bufsiz )
        buf[ x++ ] = pat[ y++ % patsiz ];
}

How about that?


Quzah.

Yes, much better.

Only if the processor is clever enough to perform the % operator using AND (meaning (patsiz & (patsiz-1)) == 0) - otherwise, I expect if (y >= patsiz) y = 0; will be faster.

--
Mats